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6.1 Graphing · PDF file 2020. 4. 6. · 535 6.1 Graphing Parabolas We already know how to graph basic parabolas which open up or down and which can be shifted around. Now, we will

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  • 535

    6.1 Graphing Parabolas

    We already know how to graph basic parabolas which open up or down

    and which can be shifted around. Now, we will generalize what we

    know in order to include parabolas which open sideways (left or right)

    and we will also be able to graph wide and narrow parabolas. We will

    utilize the process of completing the square in order to put our

    quadratics into graphing form, so you may want to review section 2.8 as

    well. We begin by comparing our basic parabola 𝑦 = π‘₯2 with the basic

    sideways parabola π‘₯ = 𝑦2. Notice that π‘₯ and 𝑦 are switched in these two equations. (This is not a coincidence and it is a property of inverse

    relations, which we study in more detail in section 6.5.)

    𝑦 = π‘₯2

    π‘₯ = 𝑦2

    π‘₯

    𝑦

    π‘₯

    𝑦 Notice that the graph of π‘₯ = 𝑦2 could also be thought of as the two graphs we

    get by solving this equation for 𝑦:

    π‘₯ = 𝑦2 implies 𝑦 = ±√π‘₯, with

    𝑦 = √π‘₯ representing the top half of the

    graph and 𝑦 = βˆ’βˆšπ‘₯

    representing the bottom half.

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    In the same way that we were able to shift the graph of 𝑦 = π‘₯2 around,

    we can also perform the same kinds of transformations on π‘₯ = 𝑦2.

    Consider the graph of 𝑦 = (π‘₯ βˆ’ 2)2 + 1, for instance. We know that one way of obtaining this graph is by first graphing our basic graph,

    𝑦 = π‘₯2, and then shifting it to the right by 2 and up 1 to obtain the following graph:

    Similarly, we could consider the graph of π‘₯ = (𝑦 βˆ’ 2)2 + 1, but to shift this, we need to keep in mind which shift is attached to which variable.

    For example, here the βˆ’2 belongs to 𝑦, and since it is on the same side

    as 𝑦, we end up going in the opposite direction (because if you solve for

    𝑦, you will end up adding 2 to the other side). Hence the graph will shift

    up 2. The value of 1 belongs to π‘₯ here, and we will end up shifting in the

    same direction since the equation is already solved for π‘₯. So the graph will shift to the right by 1:

    π‘₯

    𝑦

    π‘₯

    𝑦

    One way to remember this process

    that matches what you already know

    is to move the opposite direction for

    what is in the parentheses and the

    same direction for what is outside of

    the parentheses.

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    We will not need to spend much time on this because there is a more

    general way to graph parabolas that will include the parabolas that are

    wider and narrower that these two basic parabolas. In order to graph

    ALL parabolas, we need to allow for more than shifting and reflecting

    because there is also β€œstretching or shrinking” involved for the ones that

    have a coefficient other than 1 for the square term. We can take care of

    this β€œstretching or shrinking” by plotting a couple of points (one of these

    has to be the vertex) and using symmetry to finish our graph.

    If we are going to use this method, we must have the parabola in

    graphing form and we can put the parabola in graphing form by

    completing the square.

    For parabolas that open up or down, standard graphing form is as

    follows: π’š = 𝒂(𝒙 βˆ’ 𝒉)𝟐 + π’Œ

    From this form, we can read off the vertex (𝒉, π’Œ) which is going to be the highest or lowest point on the parabola – the place at which it β€œturns

    around”. (Note that to get the vertex, you take the opposite sign of what

    is in the parentheses and the same sign of what is outside the

    parentheses, just as when you are shifting. It is for the same reason.)

    You can also read off the axis of symmetry 𝒙 = 𝒉, which is the vertical line that divides the parabola into two equal halves. It is the line that acts

    like a mirror on one side of the parabola to give us the other side. The

    axis of symmetry is not part of the graph of the parabola, but sometimes

    it is requested in the homework problems.

    You can also tell whether the graph opens up or down by looking at the

    sign on π‘Ž. If π‘Ž is positive, then the graph opens up. If π‘Ž is negative, then the graph opens down. This really just helps you double check that you

    graphed it correctly, as plotting the points should make it open the

    correct way if there are no numerical errors.

    Once you read off the vertex, you can choose another value to plug in

    for π‘₯. (Since the domain for these types of parabolas is β€œall real

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    numbers”, you can choose any value that you like). Plug your chosen

    value into the equation to get the corresponding 𝑦 value and then plot

    your point (π‘₯, 𝑦). Then, count how far that value is away from β„Ž and if you go the same distance in the other direction, you know the y-value

    will be the same.

    The best way to make this process clear is with some examples, so first

    we will consider examples of parabolas that open up or down. Within

    the examples, we will also review the concepts of domain, range, and

    intercepts as we usually do.

    π‘₯

    𝑦

    (β„Ž, π‘˜)

    The y values should match

    for β„Ž + π‘Ÿ and β„Ž βˆ’ π‘Ÿ ,

    where π‘Ÿ is the distance you chose to go

    from β„Ž to plug in another value.

    π‘Ÿ π‘Ÿ

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    Examples

    For each of the following, graph the given relation. Give the vertex,

    axis of symmetry, domain, range, as well as the x and y intercept(s).

    1. 𝑦 = 3(π‘₯ βˆ’ 1)2 βˆ’ 2

    First, you will notice that this example is already in standard

    graphing form, so the square has already been completed. Recall

    that standard graphing form is π’š = 𝒂(𝒙 βˆ’ 𝒉)𝟐 + π’Œ.

    We should be able to read off the vertex (β„Ž, π‘˜) as well as the axis

    of symmetry π‘₯ = β„Ž pretty quickly from looking at the equation:

    vertex: (1, βˆ’2)

    axis of symmetry: π‘₯ = 1

    We can start graphing by plotting the vertex:

    π‘₯

    𝑦

    Note again that we took the opposite

    sign for what was inside of the

    parentheses and the same sign for

    what was outside.

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    Now decide what you want to plug in next. We will choose to

    plug in 0 for π‘₯ (since that will also give us the y-intercept at the same time).

    Plugging in 0 for π‘₯, we get:

    𝑦 = 3(0 βˆ’ 1)2 βˆ’ 2

    = 3(βˆ’1)2 βˆ’ 2

    = 3 βˆ™ 1 βˆ’ 2

    = 3 βˆ’ 2

    = 1

    This means we have the point (0,1) on the graph as well:

    Now to get the other side of the parabola, you can use symmetry.

    Since you traveled a distance of 1 from the vertex to get to 0, you can go

    a distance of 1 in the other direction to land at 2 and you should get the

    same y-value. That means the point (2,1) will also be on the graph.

    π‘₯

    𝑦

    We can sketch half of the parabola

    using the vertex and our new point,

    although plugging in one more value

    on this side will give you a more

    accurate sketch. (e.g. if you plug in

    βˆ’1, you will get 10)

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    Now, we can use this point to guide us when we graph the other half of

    the parabola:

    So our final graph (without the exaggerated points on it) looks like this:

    We can see the domain and range pretty clearly from the graph, but we

    also know the domain is all real numbers because this a polynomial

    equation in the variable π‘₯ (we can plug in any real number for a polynomial variable and get out a real number). Following the graph

    from bottom to top, we can see that the range is [βˆ’2, ∞) since the graph

    π‘₯

    𝑦

    π‘₯

    𝑦 There is a little shortcut that some teachers

    like to show their students that allows them

    to think of the value β€²π‘Žβ€² as a kind of β€œslope”,

    but be careful not to confuse this with

    graphing a line, because they are different

    shapes!

    With that in mind, the trick is to write β€²π‘Žβ€² as

    a fraction and move up and over (in both

    directions) from the vertex to get two

    symmetric points on the parabola. In our

    case, we would move up 3 and over 1 in

    each direction to arrive at the points we

    happened to get by plugging in a value and

    using symmetry.

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    covers all of these 𝑦 values. We already have the y-intercept also since

    we plugged in 0 earlier and found the point (0,1).

    The only thing we don’t have yet are the x-intercept(s), and from the

    graph we see that we should have two of them. To find them, we just

    plug in 0 for 𝑦 and solve for π‘₯, as usual:

    0 = 3(π‘₯ βˆ’ 1)2 βˆ’ 2

    2 = 3(π‘₯ βˆ’ 1)2

     2

    3 = (π‘₯ βˆ’ 1)2

    οο€ ο€ Β±βˆš 2

    3 = π‘₯ βˆ’ 1ο€ 

    1±√ 2

    3 = π‘₯

    Now rationalize your answers:

    π‘₯ = 1 Β± √6

    3 ο€ ο€ 

    ο€ 

    or rounding them as decimals:

     π‘₯ = 1.816

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